# rational exponents. two differing answers.

This is not homework.

Example 3) (d) of section P.4, rational exponents in Algebra and Trigonometry:

$$\frac{1}{\sqrt{x^4}} = \frac{1}{x^\frac43} = x^{-4/3}$$

Completely rational. Almost absolutely. However, expected to arrive at the same answer by simplifying the denominator first:

$$\frac{1}{\sqrt{x^{4}}} = \frac{1}{\sqrt{x^{3}}\cdot\sqrt{x}} = \frac{1}{x \sqrt{x}} = \cdots er.. \frac{1}{\sqrt{x^{2}}}\qquad?$$

It seemed like a good idea at the time. Now it just seems made up. Please help. I don't know how it got there instead.

• $x^1\cdot x^\frac 13=x^{1+\frac13}=x^\frac43$. Where did the squared term come from? – abiessu Apr 27 '14 at 20:11
• @abiessu aah.. thank you. In the simplest of places. When you post as an answer, I'll select. – user8979 Apr 27 '14 at 20:18
• @abiessu as for the squared term, saw the multiply between the two terms and just went for it. – user8979 Apr 27 '14 at 20:24

$$\frac 1{\sqrt{x^4}}=\frac 1{\sqrt{x^3}\cdot\sqrt{x}}=\frac 1{x^1\cdot x^\frac 13}=\frac 1{x^{1+\frac 13}}=\frac 1{x^{\frac 43}}=x^{-\frac 43}=\left(x^{-\frac 23}\right)^2$$
$\LARGE\ \frac{1}{x^\frac{4}{3}} = \frac{1}{x^\frac{3}{3}} * \frac{1}{x^\frac{1}{3}} = \frac{1}{x^1*x^\frac{1}{3}} = \frac{1}{x^{\frac{1}{1}+\frac{1}{3}}} = \frac{1}{x^\frac{4}{3}} = x^\frac{-4}{3}$